Unit 01 notes

Volume using cylindrical shells

Review

Shells

01 Theory

Take a graph in the first quadrant of the -plane. Rotate this about the -axis. The resulting 3D body is symmetric around the axis. We can find the volume of this body by using an integral to add up the volumes of infinitesimal shells, where each shell is a thin cylinder.

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The volume of each cylindrical shell is :

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In the limit as and the number of shells becomes infinite, their total volume is given by an integral.

Volume by shells - general formula

In any concrete volume calculation, we simply interpret each factor, ‘’ and ‘’ and ‘’, and determine and in terms of the variable of integration that is set for .

Shells vs. washers

Can you see why shells are sometimes easier to use than washers?

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02 Illustration

Example - Revolution of a triangle

01 - Revolution of a triangle

A rotation-symmetric 3D body has cross section given by the region between , , , and is rotated around the -axis. Find the volume of this 3D body.

Solution
  1. & Define the cross section region.
    • Bounded above-right by .
    • Bounded below-right by .
    • ! These intersect at .
    • Bounded at left by .
  2. && Define range of integration variable.
    • Rotated around -axis, therefore use for integration variable (shells!).
    • Integral over :
  3. & Interpret .
    • Radius of shell-cylinder equals distance along :
  4. & Interpret .
    • Height of shell-cylinder equals distance from lower to upper bounding lines:
  5. & Interpret .
    • is limit of which equals here so .
  6. & Plug data in volume formula.
    • Insert data and compute integral:

Exercise - Revolution of a sinusoid

02 - Revolution of a sinusoid

Consider the region given by revolving the first hump of about the -axis. Set up an integral that gives the volume of this region using the method of shells.

Solution

Integration by substitution

[Note: this section is non-examinable. It is included for comparison to IBP.]

03 Theory

The method of -substitution is applicable when the integrand is a product, with one factor a composite whose inner function’s derivative is the other factor.

Substitution

Suppose the integral has this format, for some functions and :

Then the rule says we may convert the integral into terms of considered as a variable, like this: ⨠⨠

The technique of -substitution comes from the chain rule for derivatives:
Here we let . Thus for some .

Now, if we integrate both sides of this equation, we find:
And of course .

Full explanation of -substitution

The substitution method comes from the chain rule for derivatives. The rule simply comes from integrating on both sides of the chain rule.

  1. && Setup: functions and .
    • Let and be any functions satisfying , so is an antiderivative of .
    • Let be another function and take for its independent variable, so we can write .
  2. ! The chain rule for derivatives.
    • Using primes notation:
    • Using differentials in variables:
  3. !!! Integrate both sides of chain rule.
    • Integrate with respect to : ⨠⨠⨠⨠
  4. &&& Introduce ‘variable’ from the -format of the integral.
    • Treating as a variable, the definition of gives:
    • Set the ‘variable’ to the ‘function’ output:
    • Combining these:
  5. && Substitute for in the integrated chain rule.
    • Reverse the equality and plug in:
  6. & This is “-substitution” in final form.

Integration by parts

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04 Theory

The method of integration by parts (abbreviated IBP) is applicable when the integrand is a product for which one factor is easily integrated while the other becomes simpler when differentiated.

Integration by parts

Suppose the integral has this format, for some functions and :

Then the rule says we may convert the integral like this: ⨠⨠

This technique comes from the product rule for derivatives:

Now, if we integrate both sides of this equation, we find:
and the IBP rule follows by algebra.

Full explanation of integration by parts

  1. && Setup: functions and are established.

    • Recognize functions and in the integrand:

  2. ! Product rule for derivatives.

    • Using primes notation:

  3. !!! Integrate both sides of product rule.

    • Integrate with respect to an input variable labeled ‘’: ⨠⨠⨠⨠

    • Rearrange with algebra:

  4. & This is “integration by parts” in final form.

Addendum: definite integration by parts

  1. ! Definite version of FTC.

    • Apply FTC to :

  2. && Integrate the derivative product rule using specified bounds.

    • Perform definite integral on both sides, plug in definite FTC, then rearrange:
Choosing factors well

IBP is symmetrical. How do we know which factor to choose for and which for ?

Here is a trick: the acronym “LIATE” spells out the order of choices – to the left for and to the right for :

05 Illustration

Example - A and T factors

03 - A and T factors

Compute the integral:

Solution
  1. & Choose .
    • Set because simplifies when differentiated.
      (By the trick: is Algebraic, i.e. more “”, and is Trig, more “”.)
    • Remaining factor must be :
  2. && Compute and .
    • Derive :
    • Antiderive :
    • Obtain chart:
  3. && Plug into IBP formula.
    • Plug in all data:
    • Compute integral on RHS: Note: the point of IBP is that this integral is easier than the first one!
  4. & Final answer is:

Exercise - Hidden A

04 - Hidden A

Compute the integral:

Solution

Trig power products

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06 Theory

Review: trig identities

Trig power product:

A power product has this form:

for some integers and (even negative!).

To compute these integrals, use a sequence of these techniques:

  • Swap an even bunch.
  • -sub for power-one.
  • Power-to-frequency conversion.
  • Memorize these three techniques!

Examples of trig power products:

Swap an even bunch

If either or is an odd power, use ⨠⨠⨠⨠ (maybe repeatedly) to convert an even bunch to the opposite trig type.

An even bunch is all but one from the odd power.

For example: ⨠⨠⨠⨠⨠⨠⨠⨠⨠⨠

-sub for power-one

If or , perform -substitution to do the integral.

The other trig power becomes a power; the power-one becomes .

For example, using and thus we can do:
⨠⨠⨠⨠

  • ! By combining these tricks you can do any power product with at least one odd power!
    • Leave a power-one from the odd power when swapping an even bunch.
  • !! Notice: , even powers. So the method works for and similar.
Power-to-frequency conversion

Using these ‘power-to-frequency’ identities (maybe repeatedly):

change an even power (either type) into an odd power of cosine.

For example, consider the power product:
You can substitute appropriate powers of and :
⨠⨠⨠⨠By doing some annoying algebra, this expression can be expanded as a sum of smaller powers of :
⨠⨠

Each of these terms can be integrated by repeating the same techniques.

07 Illustration

Example - Trig power product with an odd power

05 - Power product: odd power

Compute the integral:

Solution
  1. &&& Swap over the even bunch.
    • Max even bunch leaving power-one is : ⨠⨠⨠⨠
    • Apply to in the integrand: ⨠⨠
  2. &&& Perform -substitution on the power-one integrand.
    • Set .
    • Hence . Recognize this in the integrand.
    • Convert the integrand: ⨠⨠⨠⨠
  3. & Perform the integral.
    • Expand integrand and use power rule to obtain:
    • Insert definition : ⨠⨠⨠⨠
  4. & This is our final answer.

08 Theory

Trig power product: or

A power product has this form: A power product has this form:

To integrate these, swap an even bunch using:

OR:

Or do -substitution using:

OR:

Note:

  • There is no simple “power-to-frequency conversion” for tan / sec !

We can modify the power-one technique to solve some of these. We need to swap over an even bunch from the odd power so that exactly the factor is left behind.

Considering all the possibilities, one sees that this method works when:

  • is an odd power
  • is an even power

Quite a few cases escape this method:

  • Any
  • Any for even and odd

These tricks don’t work for or or , among others.

Special integrals: tan and sec

We have:

  • These integrals should be memorized individually.
Deriving special integrals - tan and sec

The first formula can be found by -substitution, considering that .

The second formula can be derived by multiplying by a special “”, computing instead by expanding the numerator and doing -sub on the denominator.

09 Illustration

Example - Trig power product with tan and sec

06 - Power product: tan and sec

Compute the integral:

Solution
  1. && Try .
    • Factor out of the integrand: ⨠⨠
    • We then must swap over remaining into the type.
    • Cannot do this because has odd power. Need even to swap.
  2. && Try .
    • Factor out of the integrand: ⨠⨠
    • Swap remaining into type: ⨠⨠
    • Substitute and : ⨠⨠
  3. &&& Compute the integral in and convert back to .
    • Expand the integrand: ⨠⨠
    • Apply power rule: ⨠⨠
    • Plug back in, : ⨠⨠
  4. & This is our final answer.

Trig substitution

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10 Theory

Certain algebraic expressions have a secret meaning that comes from the Pythagorean Theorem. This meaning has a very simple expression in terms of trig functions of a certain angle.

For example, consider the integral: Now consider this triangle:

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The triangle determines the relation , and it implies .

Now plug these into the integrand above: ⨠⨠Considering that , we obtain a very reasonable trig integral:
⨠⨠⨠⨠⨠⨠We must rewrite this in terms of using to finish the problem. We need to find assuming that . To do this, refer back to the triangle to see that . Plug this in for our final value of the integral: ⨠⨠

Here is the moral of the story:

  • Re-express the Pythagorean expression using a triangle and a trig substitution.
    • In this way, square roots of quadratic polynomials can be eliminated.

There are always three steps for these trig sub problems:

  • (1) Identify the trig sub: find the sides of a triangle and relevant angle .
  • (2) Solve a trig integral (often a power product).
  • (3) Refer back to the triangle to convert the answer back to .

To speed up your solution process for these problems, memorize these three transformations:

(1)
⨠⨠(2)
⨠⨠(3)
⨠⨠

For a more complex quadratic with linear and constant terms, you will need to first complete the square for the quadratic and then do the trig substitution.

11 Illustration

Example - Trig sub in quadratic: completing the square

08 - Trig sub in quadratic: completing the square

Compute the integral:

Solution
  1. ! Notice square root of a quadratic.
  2. &&& Complete the square to obtain Pythagorean form.
    • Find constant term for a complete square:
    • Add and subtract desired constant term: ⨠⨠
    • Simplify: ⨠⨠
  3. && Perform shift substitution.
    • Set as inside the square:
    • Infer .
    • Plug into integrand: ⨠⨠
  4. !! Trig sub with .
    • Identify triangle:
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    • Use substitution . (From triangle or memorized tip.)
    • Infer .
    • Plug in data: ⨠⨠
  5. & Compute trig integral.
    • Use ad hoc formula:
  6. && Convert trig back to .
    • First in terms of , referring to the triangle:
    • Then in terms of using .
    • Plug everything in: ⨠⨠
  7. && Simplify using log rules.
    • Log rule for division gives us:
    • The common denominator can be pulled outside as .
    • The new term can be “absorbed into the constant” (redefine ).
    • So we write our final answer thus: